problem Problem 2

Internal problem ID [2885]
Internal ﬁle name [OUTPUT/2377_Sunday_June_05_2022_03_02_43_AM_57791878/index.tex]

Book: Diﬀerential equations and linear algebra, Stephen W. Goode and Scott A Annin. Fourth edition, 2015
Section: Chapter 10, The Laplace Transform and Some Elementary Applications. Exercises for 10.8. page 710
Problem number: Problem 2.
ODE order: 1.
ODE degree: 1.

The type(s) of ODE detected by this program : "exact", "linear", "ﬁrst_order_ode_lie_symmetry_lookup"

\[ \boxed {y^{\prime }-2 y=\delta \left (t -2\right )} \] With initial conditions \begin {align*} [y \left (0\right ) = 1] \end {align*}

15.2.1 Existence and uniqueness analysis

This is a linear ODE. In canonical form it is written as \begin {align*} y^{\prime } + p(t)y &= q(t) \end {align*}

Where here \begin {align*} p(t) &=-2\\ q(t) &=\delta \left (t -2\right ) \end {align*}

Hence the ode is \begin {align*} y^{\prime }-2 y = \delta \left (t -2\right ) \end {align*}

15.2.2 Solving as laplace ode

Solving using the Laplace transform method. Let \[ \mathcal {L}\left (y\right ) =Y(s) \] Taking the Laplace transform of the ode and using the relations that \begin {align*} \mathcal {L}\left (y^{\prime }\right )&= s Y(s) - y \left (0\right ) \end {align*}

The given ode now becomes an algebraic equation in the Laplace domain \begin {align*} s Y \left (s \right )-y \left (0\right )-2 Y \left (s \right ) = {\mathrm e}^{-2 s}\tag {1} \end {align*}

Replacing initial condition gives \begin {align*} s Y \left (s \right )-1-2 Y \left (s \right ) = {\mathrm e}^{-2 s} \end {align*}

Solving for \(Y(s)\) gives \begin {align*} Y(s) = \frac {{\mathrm e}^{-2 s}+1}{s -2} \end {align*}

Taking the inverse Laplace transform gives \begin {align*} y&= \mathcal {L}^{-1}\left (Y(s)\right )\\ &= \mathcal {L}^{-1}\left (\frac {{\mathrm e}^{-2 s}+1}{s -2}\right )\\ &= {\mathrm e}^{2 t}+{\mathrm e}^{2 t -4} \left (-\operatorname {Heaviside}\left (-t +2\right )+1\right ) \end {align*}

Converting the above solution to piecewise it becomes \[ y = \left \{\begin {array}{cc} {\mathrm e}^{2 t} & t \le 2 \\ {\mathrm e}^{2 t}+{\mathrm e}^{2 t -4} & 2

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= \left \{\begin {array}{cc} {\mathrm e}^{2 t} & t \le 2 \\ {\mathrm e}^{2 t}+{\mathrm e}^{2 t -4} & 2

Veriﬁcation of solutions

\[ y = \left \{\begin {array}{cc} {\mathrm e}^{2 t} & t \le 2 \\ {\mathrm e}^{2 t}+{\mathrm e}^{2 t -4} & 2

15.2.3 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime }-2 y=\mathit {Dirac}\left (t -2\right ), y \left (0\right )=1\right ] \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=2 y+\mathit {Dirac}\left (t -2\right ) \\ \bullet & {} & \textrm {Group terms with}\hspace {3pt} y\hspace {3pt}\textrm {on the lhs of the ODE and the rest on the rhs of the ODE}\hspace {3pt} \\ {} & {} & y^{\prime }-2 y=\mathit {Dirac}\left (t -2\right ) \\ \bullet & {} & \textrm {The ODE is linear; multiply by an integrating factor}\hspace {3pt} \mu \left (t \right ) \\ {} & {} & \mu \left (t \right ) \left (y^{\prime }-2 y\right )=\mu \left (t \right ) \mathit {Dirac}\left (t -2\right ) \\ \bullet & {} & \textrm {Assume the lhs of the ODE is the total derivative}\hspace {3pt} \frac {d}{d t}\left (y \mu \left (t \right )\right ) \\ {} & {} & \mu \left (t \right ) \left (y^{\prime }-2 y\right )=y^{\prime } \mu \left (t \right )+y \mu ^{\prime }\left (t \right ) \\ \bullet & {} & \textrm {Isolate}\hspace {3pt} \mu ^{\prime }\left (t \right ) \\ {} & {} & \mu ^{\prime }\left (t \right )=-2 \mu \left (t \right ) \\ \bullet & {} & \textrm {Solve to find the integrating factor}\hspace {3pt} \\ {} & {} & \mu \left (t \right )={\mathrm e}^{-2 t} \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} t \\ {} & {} & \int \left (\frac {d}{d t}\left (y \mu \left (t \right )\right )\right )d t =\int \mu \left (t \right ) \mathit {Dirac}\left (t -2\right )d t +c_{1} \\ \bullet & {} & \textrm {Evaluate the integral on the lhs}\hspace {3pt} \\ {} & {} & y \mu \left (t \right )=\int \mu \left (t \right ) \mathit {Dirac}\left (t -2\right )d t +c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y=\frac {\int \mu \left (t \right ) \mathit {Dirac}\left (t -2\right )d t +c_{1}}{\mu \left (t \right )} \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} \mu \left (t \right )={\mathrm e}^{-2 t} \\ {} & {} & y=\frac {\int {\mathrm e}^{-2 t} \mathit {Dirac}\left (t -2\right )d t +c_{1}}{{\mathrm e}^{-2 t}} \\ \bullet & {} & \textrm {Evaluate the integrals on the rhs}\hspace {3pt} \\ {} & {} & y=\frac {\mathit {Heaviside}\left (t -2\right ) {\mathrm e}^{-4}+c_{1}}{{\mathrm e}^{-2 t}} \\ \bullet & {} & \textrm {Simplify}\hspace {3pt} \\ {} & {} & y={\mathrm e}^{2 t} \left (\mathit {Heaviside}\left (t -2\right ) {\mathrm e}^{-4}+c_{1} \right ) \\ \bullet & {} & \textrm {Use initial condition}\hspace {3pt} y \left (0\right )=1 \\ {} & {} & 1=c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} c_{1} \\ {} & {} & c_{1} =1 \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} c_{1} =1\hspace {3pt}\textrm {into general solution and simplify}\hspace {3pt} \\ {} & {} & y={\mathrm e}^{2 t} \left (\mathit {Heaviside}\left (t -2\right ) {\mathrm e}^{-4}+1\right ) \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y={\mathrm e}^{2 t} \left (\mathit {Heaviside}\left (t -2\right ) {\mathrm e}^{-4}+1\right ) \end {array} \]

Maple trace

`Methods for first order ODEs: 
--- Trying classification methods --- 
trying a quadrature 
trying 1st order linear 
<- 1st order linear successful`

\[ y \left (t \right ) = \operatorname {Heaviside}\left (t -2\right ) {\mathrm e}^{-4+2 t}+{\mathrm e}^{2 t} \]

15.2 problem Problem 2

15.2.1 Existence and uniqueness analysis

15.2.2 Solving as laplace ode

15.2.3 Maple step by step solution